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Inverse function containing square root

  1. Sep 13, 2009 #1
    1. The problem statement, all variables and given/known data

    Given f(x) = -x2 - 2x + 3, x < -1

    Find f-1.
    Find f-1f.

    2. Relevant equations

    Nil.

    3. The attempt at a solution

    Actually I worked out much of the question already, and I already know that f-1f: x -> x. The problem is, I can't seem to get f-1f(x) = x on substitution. Here goes:

    Let y = -x2 - 2x + 3
    4 - y = (x+1)2
    x = - 1 +/- sqrt(4-y)

    Since the domain of f is the range of f-1, we have

    f-1: x -> -1 - sqrt(4-x), x < 4

    Consider f-1(x) = -1 - sqrt(4-x)
    Substituting f(x),
    f-1(f(x)) = -1 - sqrt(x2+2x+1) = -x-2

    OK, here's the problem. I know I will get f-1f(x)= x if sqrt(x2+2x+1) = -x-1 and I understand that

    x2+2x+1 = (x+1)2 = (-x-1)2

    but my reasoning here doesn't reject -x-2 as a valid answer. Where is the mistake in my working?
     
  2. jcsd
  3. Sep 13, 2009 #2
    This is not correct.
    [tex]-1 - \sqrt{x^2 + 2x + 1} = -1 - \sqrt{(x+1)^2} = -1 - |x + 1|[/tex]
    Since the domain of [itex]f^{-1}\circ f[/itex] is the same as the domain of f, giving only negative values for the expression in the absolute value function, there is only one branch of the absolute value function in use, which should allow you to derive the correct identity.
     
    Last edited: Sep 13, 2009
  4. Sep 13, 2009 #3
    Ahhh I totally forgot about the modulus. Thanks a lot!
     
  5. Sep 15, 2009 #4

    Mark44

    Staff: Mentor

    Another point that should be mentioned has to do with your notation. f-1f means the product of the inverse and the original function. That's different from the composition of the two functions, which is represented this way:

    [itex]f^{-1}\circ f[/itex]

    The composition of these two functions is defined this way:
    [itex](f^{-1}\circ f)(x)[/itex] = f-1(f(x))
     
  6. Sep 15, 2009 #5
    Actually, I've seen the circle notation in some texts and product notation in other texts to reflect their relation to real number multiplication as is done in abstract algebra. The circle notation is not required.
     
  7. Sep 16, 2009 #6

    Mark44

    Staff: Mentor

    In abstract algebra you have structures such as groups in which you have a set of objects and an operation. The operation can be normal addition or multiplication, or can be function composition. In any case, the operation is well-understood.

    If you're working with functions gf usually means the product of two functions, while g[itex]\circ[/itex] f normally means function composition. To avoid confusing the two operations, a small circle is usually used. That was my point.
     
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